Consider the following reaction at 25^∘ C : NO(g)+1 / 2 O2( g) ⟶NO2( g) for which ΔH^∘=-57.0 kJ

Consider the following reaction at 25^∘ C : NO(g)+1 / 2...

Consider the following reaction at $25^{\circ} \mathrm{C}$ : $$ \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{NO}_2(\mathrm{~g}) $$ for which $\Delta H^{\circ}=-57.0 \mathrm{~kJ}$ and $K=1.50 \times 10^6$. Calculate $\Delta \varsigma^{\circ}$ for the following reaction: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g}) \quad \Delta S^{\circ}=\text { ? } $$

Consider the following reaction at $25^{\circ} \mathrm{C}$ :
$$
\mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{NO}_2(\mathrm{~g})
$$
for which $\Delta H^{\circ}=-57.0 \mathrm{~kJ}$ and $K=1.50 \times 10^6$. Calculate $\Delta \varsigma^{\circ}$ for the following reaction:
$$
2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g}) \quad \Delta S^{\circ}=\text { ? }
$$

Key Concepts

Entropy Change

Entropy change (?S°) in a chemical reaction measures the change in disorder or randomness as reactants are converted into products. By utilizing the equation ?G° = ?H° - T?S° along with known values of ?H° and the equilibrium constant (which gives ?G°), the standard entropy change can be computed, reflecting the inherent dispersal of energy in the process.

Reaction Stoichiometry Scaling

When a reaction is scaled by a stoichiometric coefficient, the associated thermodynamic quantities such as ?H°, ?S°, and ?G° are also scaled by the same factor. This principle, akin to Hess's law, ensures that multiplying a balanced reaction by a number properly modifies its enthalpy and entropy changes.

Gibbs Free Energy

Gibbs free energy, defined as ?G° = ?H° - T?S°, is a central concept in thermodynamics that determines the spontaneity of a reaction. It provides a link between enthalpy (heat content), entropy (degree of disorder), and temperature, allowing one to calculate any one of these parameters if the others are known.

Equilibrium Constant and Free Energy Relationship

The equilibrium constant (K) of a reaction is directly related to the standard free energy change by the relationship ?G° = -RT ln K. This connection allows the conversion of experimental equilibrium data into thermodynamic quantities, providing insight into the driving forces governing the reaction.

Transcript

00:01 So for this question, we're trying to calculate the value of k where the temperature is 25 degrees celsius.

00:07 So first we're going to use the enthalpy, entropy, and temperature values to calculate the gibbs free energy using the equation where the change in gibbs free energy is equal to the change in enthalpy minus the temperature times the change in entropy.

00:22 We're going to use that calculation or that equation to calculate the gibbs free energy and then once we have the gibbs free energy, we can use the equation where the gibbs change in gibbs free energy is equal to negative r, which is a constant, times the temperature times the natural log of k to solve for k.

00:45 So first let's figure out the gibbs free energy at 25 degrees celsius.

00:50 So before we can plug these into our equation, we have to make sure the units match.

00:54 So because the entropy is in kelvin, let's convert our temperature into kelvin, which we just add 273.

01:01 Which gives us 298 kelvin.

01:06 And then our enthalpy is in kilojoules and our entropy is in joules.

01:13 So we'll turn the joules into kilojoules by dividing by a thousand, which gives us negative 0 .1766 kilojoules over kelvin times moles.

01:30 So now we can plug these values in to solve for the change in gibbs free energy.

01:35 So it's equal to the change in enthalpy, negative 58 .03 minus the temperature, 298 kelvin, times the change in entropy, which is negative 0 .1766.

01:52 Type that into your calculator and you should get negative 5 .40 kilojoules.

02:03 So now that we have our gibbs free energy, we can plug that into the second equation to solve for k.

02:15 So our gibbs free energy, negative 5 .40 kilojoules per mole, is equal to negative r, which r is in, i forgot, r is in joules.

02:30 The value for r is 8 .314 joules per kelvin times mole.

02:45 So our gibbs free energy needs to be in joules, so we just take that times a thousand, which gives us negative 5400 joules per mole.

03:03 So let's change that in our equation.

03:07 So negative 5400 joules per mole is equal to negative r, which is 8 .314, times the temperature, which is 298 kelvin, times the natural log of k...

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